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18

Hint: $\pi$ is not a Liouville number, so there exists $m\in\mathbb{N}$ such that for all $p,q\in\mathbb{Z}$ with $q>1$, we have $$ \left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$ This should allow you to keep $\sin n$ away from 0. Edit: Full Solution: Let $m$ be as above. So for all $p,q\in\mathbb{Z}$ with $q>1$ we have $$ \left| \pi - ...


17

Supose not, so that there exists an $\varepsilon>0$ such that $(0,\varepsilon)\cap S=\emptyset$. $\qquad\qquad\qquad(\star)$ It follows that $\alpha=\inf S\cap(0,+\infty)$ is a positive number. The choice of $\alpha$ and its positivity implies that the one and only element of $S$ which is in $[0,\alpha)$ is $0$. I claim that $\alpha\in S$. ...


15

This isn't exceptionally good compared to the partial convergents of the continued fraction expansion. Terminating the continued fraction for $\pi$ right before the $292$ gives $\frac{355}{113}= \textbf{3.141592}9203\ldots$, which gets $6$ digits after the decimal place right, while your fraction only gets $3$ digits after the decimal place correct even ...


10

Though it is clear what would count as a positive answer, the question is not precise enough that to make clear what counts as a negative answer. Nonetheless, I believe that no similar formula could possibly be equivalent to denseness. The problem is that there are sequences that are dense but that are wildly non-equidistributed. For example, your formula ...


10

Using the property stated in that article: $$\mu(x)=2 + \limsup \frac{\log a_{n+1}}{\log q_n}$$ where the continued fraction expansion for $x$ is $[a_0,a_1,...]$ and the $n$th convergent is $\frac{p_n}{q_n}$. Start with $a_0=2$ and $a_1=2$, so $q_0=1$, $q_1=2$. Now, assume you have a continued fraction $$\frac{p_n}{q_n}=[a_0,...,a_n]$$ Define $a_{n+1}$ ...


9

Actually, we can do it for higher powers as well. Let $F_k (N)$ denote the sum of the $k^{th}$ powers of the differences in the Farey sequence. Then it is easy to see that $$F_0(N)=\sum_{n=1}^N \phi(n)\sim \frac{3N^2}{\pi^2}$$ and $$F_1(N)=1.$$ It seems you are looking for $F_2 (N)$, and, you are correct that the asymptotic is $\frac{\log N}{N^2}$. More ...


8

There are some details that I haven't fully vetted but here's a long sketch which I believe should show divergence. Define $c_n := \frac{n}{2\pi} \pmod 1$ and let $[a,b]$ be any interval in the torus $\mathbb R/\mathbb Z$. The discrepancy $D(N)$ of the sequence $c_n$ is defined to be the difference between $\# \{ n \le N : c_n \in [a,b]\}$ and the expected ...


8

Suppose that the prime factors of $n^2+1$ are all bounded by $N$ for infinitely many $n$. Then infinitely many integers $n^2 + 1$ can be written in the form $D y^3$ for one of finitely many $D$. Explicitly, the set of $D$ can be taken to be the finitely many integers whose prime divisors are all less than $N$, and whose exponents are at most $2$. For ...


7

I proceed the same way as in my other answer here http://math.stackexchange.com/a/110019/17445 $|\sin n| \le \varepsilon$ implies that there exists an integer $k(n)$ such that $n = k(n)\pi + a(n)$ where $a(n) \in [-\arcsin(\varepsilon) ; \arcsin(\varepsilon)] \subset [-\pi \varepsilon/2 ; \pi \varepsilon/2]$, and if both $|\sin n|$ and $|\sin m|$ are less ...


7

Let $x$ be a fixed positive real. Then, the behavior of the sequence $\{ kx \}$ falls under one of the following three types, depending on $x$. If $x$ is an integer: In this case, the sequence $\{ kx \}$ is the constant sequence $0$. In particular, the sequence converges to $0$. If $x$ is rational, but not an integer: Suppose $x = a/b$ where $a,b$ are ...


7

Assume that $a, b \in \mathbb N $ are not perfect squares and $ \sqrt a + \sqrt b = n \in \mathbb N$. Then $$ a = (n - \sqrt b)^2 = n^2 - 2 n \sqrt b + b $$ which means that $\sqrt b$ is a rational number. This contradicts the fact that the square root of an integer is either an integer or a irrational number.


7

The problem is that equidistribution is a property of sequences, but density is a property of sets. You may wish to prove as an exercise that a countable subset of $[0,1]$ is dense if and only if it can be ordered in such a way as to be equidistributed. So the set $\lbrace\,x_0,x_1,\dots\,\rbrace$ is dense if and only if there is a permutation $\sigma$ such ...


7

normal $\; \not\Rightarrow \;$ irrationality measure $2$ There exist numbers that are normal with irrationality measure $>2$. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure $\infty.$ This is Theorem 2 in Bugeaud [2] (2002). For related results, see [1] and [3]. irrationality measure $2$ ...


7

Suppose $a+b=c$, so that $a+b-c=0$, with $a^3, b^3, c$ all rational. Then we have $-3abc=a^3+b^3-c^3$ by virtue of the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ (take $c$ with a negative sign) Hence $a+b$ and $ab$ are both rational, so $a$ and $b$ satisfy a quadratic equation with rational coefficients. There are lots of ways of ...


6

You can determine this by looking at the continued fractions of 0.8575 and 0.8585. They are [0, 1, 6, 57] and [0, 1, 6, 14, 1, 8, 2] so anything in [0, 1, 6, 15..57] would work. [0, 1, 6, 15] has the smallest denominator in that range which corresponds to 0+1/(1+1/(6+1/15)) = 91/106. I just tested smaller denominators and this checks out right. ...


5

If we look at the continued fraction of the expression $\frac{\pi}{180}$, we get: $$\frac\pi{180} = [0,57,3,2,1,1,1,2,40,\dots]$$ And: $$\frac{44}{2521} = [0,57,3,2,1,1,1,1]$$ So that's a pretty good match. It is between two terms of the continued fraction expansion of $\frac\pi{180}$, in particular. Now, whether this is surprisingly good in any way, ...


5

Taking a step into generalization, it is true that every additive subgroup $G$ of $\mathbb R$ is either discrete or dense. This can be proved by considering $\alpha = \inf \{ x \in G : x>0 \}$. Then $G$ is discrete iff $\alpha >0$, in which case $G=\alpha \mathbb Z$. In your case, Jyrki's suggestion implies that $\alpha=0$ and so $S$ is dense.


5

The Weyl equidistribution theorem says that for irrational $\alpha$ and sufficiently many $k$, the fractional parts $\{\alpha k\}$ will be equidistributed in the interval $[0,1]$. To apply this to your particular problem, consider a large interval $k\in[N,2N]$. Because of equidistribution, the numbers $k$ will "hit" the condition $\{\alpha k\} < ...


4

Some 'brute force' solutions of $\ x^3 + x + y^3 + y - z^3 - z = r$ (with additional contributions from Oleg567) For $\ r=0\ $ (and $z<10^5$) : \begin{array} {cc|c} x&y&z\\ \hline 36& 37& 46\\ 98& 248& 253\\ 165& 705& 708\\ 320& 377& 442\\ 843& 1078& 1228\\ 2372& 3323& 3685\\ 2988& 3070& ...


4

A good way to construct fractions which approximate numbers is to use continued fractions. Namely, if we have that $x=a_0+1/(a_1+1/(a_2+1/(\cdots$ (which we shall write with the shorthand $x=[a_0,a_1,a_2,\ldots]$), we can define the convergents of $x$ to be the partial sums $[a_1, \ldots, a_k]=a_0+1/(a_1 + 1/(\ldots +1/(a_k))\cdots))$. These will give the ...


3

Let $N=8$, $k=2$, $a_1=1$, $a_2=5$. Then $a_1x/N=x/8$ and its distance from the nearest integer is $1/8$ if $x$ is $8r\pm1$, $3/8\gt1/4$ if $x$ is $8r\pm3$; $a_2x/N=5x/8$ and its distance from the nearest integer is $1/8$ if $x$ is $8r\pm3$, $3/8\gt1/4$ if $x$ is $8r\pm1$. So there is no $x$, $\gcd(x,N)=1$, such that both $a_1x/N$ and ...


3

If $\left|\frac pq-r\right|\lt\frac1{2q^2}$, then $\frac pq$ is a convergent for the continued fraction for $r\not\in\mathbb{Q}$. If $\frac pq$ is a convergent for the continued fraction for $r\not\in\mathbb{Q}$, then $\left|\frac pq-r\right|\lt\frac1{q^2}$ Thus, I would call $\frac pq$ is a "good" approximation for $r$ if $\left|\frac ...


3

More generally, for $n\in\mathbb N$ let $a_1,\ldots,a_n\in\mathbb N$ be distinct and square-free and let $c_1,\ldots,c_n\in\mathbb Q$. Then the number $s=\sum_{k=1}^nc_k\sqrt{a_k} $ is rational if and only if $c_1=\ldots=c_n=0$. In other words, even something like $\sqrt n+\sqrt m-\sqrt k\in\mathbb Q$ has only "trivial" solutions such as $\sqrt 2+\sqrt ...


3

The proof for solutions of $\displaystyle{2x^2+1=3^n}$ can be read from the paper at American Mathematical Society Volume 131, Number 12 According to that three solutions are $(1,1), (2,2)$ and $(11,5)$ NOTE: I believe one cannot attempt with one approach to solve all of those equations. ADDING THESE NOTE (Since it was requested in the comment here) A ...


3

A proof of the fact that the set of reals with irrationality measure $\gt 2$ has Lebesgue measure $0$ can be found in several places. I think the result is due to Khinchin, and is in his book on continued fractions. Here is an online proof by Beukers. It is towards the end of the paper, but independent of the rest, which anyway is worth reading. It ...


3

Here are some necessary and sufficient conditions: 1.a. The set $\{n\mid a\leqslant x_n\leqslant b\}$ is nonempty for every $0\leqslant a<b\leqslant 1$. 1.b. The set $\{n\mid a\leqslant x_n\leqslant b\}$ is infinite for every $0\leqslant a<b\leqslant 1$. 2.a. The sum of the series $\sum\limits_nf(x_n)$ is positive for every continuous nonnegative ...


3

If you know how to solve problems of the form "find the least integer $n(\eta) \ge 1$ such that $n\alpha \in [- \eta - \varepsilon; - \eta + \varepsilon] \pmod 1$" where $|\eta| \le \varepsilon$, then you can easily find all those integers one after the other. Thus, you want to keep a list of the integers $n$ such that $n \alpha$ is closer to $0$ than for ...


3

The statement can be false even when you assume $\frac{\theta}{\pi}$ is irrational. Let $(s_k)_{k=1,2\ldots}$ be the sequence defined recursively by: $$s_k = \begin{cases}1,&\quad\text{ for }k = 1\\2^{s_{k-1}},&\quad\text{ for }k > 1\end{cases}$$ and $s$ be the Liouville number $s = \sum_{k=1}^{\infty} 2^{-s_k}$. For $r = \sqrt{2}$, and $\theta ...


3

There's a fairly "dumb" but surprisingly effective trick here: every number of the form $3^n$ is either of the form $x^3$, $3x^3$, or $9x^3$. This idea allows you to map exponential Diophantine equations into finitely many polynomial ones: the new equations are so much less restrictive, but generally better understood.



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